Wall with openings: stress, brick, mortar

Data and assumptions

• Wall 230 mm thick, modular bricks ($k_p=1$). Total length 5.26 m ($230+2000+1000+1800+230$ mm), with a 1.0 m wide opening up to lintel level 2.0 m. Solid length above plinth $=5.26-1.00=4.26$ m; below plinth level the wall is continuous over the full 5.26 m.
• Ceiling height 2.8 m; 115 mm one-way RCC slab of 3.5 m clear span bearing on this wall. Finish and roofing assumed 1.0 kN/m$^2$ (not given). Live load 1.5 kN/m$^2$. Eccentricity of the slab load assumed $e/t=1/12$ (IS 1905 App. A-2).
• The load over the opening is carried by the lintel to the solid wall on either side (IS 1905 cl. 5.3.2, 5.3.3), so above the plinth the solid parts carry the whole load.
• Two sections are checked: just above plinth level (solid wall only) and at the top of the footing (full length, including the plinth wall).
• Brick masonry 19 kN/m$^3$, RCC 25 kN/m$^3$.

Step 1: Loads on the wall

$$\begin{aligned} w &= 25\times0.115+1.0+1.5 = 5.375\ \text{kN/m}^2\\ \text{Slab reaction per m} &= w\times\frac{3.5}{2} = 9.41\ \text{kN/m}\\ \text{Slab load on whole wall} &= 9.41\times5.26 = 49.5\ \text{kN}\\ \text{Wall above lintel (over opening)} &= 0.23\times1.0\times(2.8-2.0)\times19 = 3.50\ \text{kN}\\ \text{Solid wall self weight} &= 0.23\times4.26\times2.8\times19 = 52.1\ \text{kN}\\ N_{plinth} &= 105.1\ \text{kN}\\ \text{Plinth wall (full length)} &= 0.23\times5.26\times1.5\times19 = 34.5\ \text{kN}\\ N_{footing} &= 105.1+34.5 = 139.6\ \text{kN} \end{aligned}$$

Step 2: Maximum stress in the wall

$$\begin{aligned} \text{Above plinth: } \sigma &= \frac{105.1\times10^3}{0.23\times4.26\times10^6} = 0.107\ \text{N/mm}^2\\ \text{At top of footing: } \sigma &= \frac{139.6\times10^3}{0.23\times5.26\times10^6} = 0.115\ \text{N/mm}^2 \end{aligned}$$

Bending from the slab eccentricity ($e = t/12 = 19.2$ mm on the 49.5 kN slab load): $M = 49.5\times0.0192 = 0.95$ kNm. At the footing, $Z = 5260\times230^2/6 = 46.4\times10^6$ mm$^3$, so $\sigma_b = 0.95\times10^6/46.4\times10^6 = 0.020$ N/mm$^2$.

Maximum stress in the wall: 0.115 N/mm$^2$ (average) at the top of the footing, or about $0.115+0.020 = 0.136$ N/mm$^2$ at the extreme fibre. (Just above plinth level: 0.107 average, about 0.132 at the extreme fibre.)

Step 3: Slenderness and permissible stress

Above plinth: the solid parts (2.23 m and 2.03 m long) are longer than $4t = 0.92$ m, so they act as walls, not as columns between openings. With the RCC slab at the top and the plinth floor at the bottom (full restraint), $H=2.8+0.115/2=2.858$ m:

$$\begin{aligned} h_{ef} &= 0.75\times2.858 = 2.143\ \text{m}\\ SR &= \frac{2.143}{0.23} = 9.32\ (<27) \end{aligned}$$

$k_s=0.894$ (Table 9, $e/t=1/12$), $k_a=1.0$ (section area greater than 0.2 m$^2$), $k_p=1$, and a 25 percent increase applies ($e/t>1/24$):

$$f_b \ge \frac{0.107}{0.894\times1.25} = 0.096\ \text{N/mm}^2$$

At the top of the footing (full-length wall, conservatively taking the full height from footing to slab): $H = 1.5+2.8+0.0575 = 4.36$ m, $h_{ef}=0.75\times4.36=3.27$ m, $SR=14.2$ ($<27$), $k_s = 0.734$:

$$f_b \ge \frac{0.115}{0.734\times1.25} = 0.126\ \text{N/mm}^2$$

The section at the top of the footing governs (required $f_b = 0.126$ N/mm$^2$).

Step 4: Brick and mortar (IS 1905 Table 8)

Answer: maximum stress about 0.115 N/mm$^2$ (average; about 0.136 N/mm$^2$ at the extreme fibre) at the top of the footing; required $f_b = 0.126$ N/mm$^2$, so the lowest class of brick (3.5 N/mm$^2$) with M2 mortar (1:6, $f_b=0.35$) is adequate. For durability and practical reasons, adopt bricks of not less than 5 N/mm$^2$ with 1:6 cement mortar. Provide lintels with a bearing of at least 230 mm (or 1/10 of the span) on each side.